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If A + B + C = \(\frac{\pi}{2}\). P.T tanA · tanB + tanB . tanC + tanC · tanA = 1 |
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Answer» Given A + B + C = \(\frac{\pi}{2}\); A + B = 90° – C; tan(A + B) = tan(90° – C) \(\frac{tanA + tanB}{1 - tanAtanB} = cotc \) \(\frac{1}{tanC}\) ⇒ tanA tanC + tanB tanC = 1 – tanA · tanB ⇒ tan A tanB + tanB tanC + tanC tanA = 1. |
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